Precalculus Examples

$\ln (x^{2} + x - 20) = 1$

Step 1

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x^{2} + x - 20)} = e^{1}$

Step 2

Rewrite $\ln (x^{2} + x - 20) = 1$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{1} = x^{2} + x - 20$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} + x - 20 = e^{1}$ .

$x^{2} + x - 20 = e$

Step 3.2

Subtract $e$ from both sides of the equation.

$x^{2} + x - 20 - e = 0$

Step 3.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.4

Substitute the values $a = 1$ , $b = 1$ , and $c = - 20 - e$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot (- 20 - e))}}{2 \cdot 1}$

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot (- 20 - e)}}{2 \cdot 1}$

Step 3.5.1.2

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot (- 20 - e)}}{2 \cdot 1}$

Step 3.5.1.3

Apply the distributive property.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 20 - 4 (- e)}}{2 \cdot 1}$

Step 3.5.1.4

Multiply $- 4$ by $- 20$ .

$x = \frac{- 1 \pm \sqrt{1 + 80 - 4 (- e)}}{2 \cdot 1}$

Step 3.5.1.5

Multiply $- 1$ by $- 4$ .

$x = \frac{- 1 \pm \sqrt{1 + 80 + 4 e}}{2 \cdot 1}$

Step 3.5.1.6

Add $1$ and $80$ .

$x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2 \cdot 1}$

Step 3.5.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{81 + 4 e}}{2}$

Step 3.6

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{81 + 4 e}}{2}, - \frac{1 + \sqrt{81 + 4 e}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{81 + 4 e}}{2}, - \frac{1 + \sqrt{81 + 4 e}}{2}$

Decimal Form:

$x = 4.29252353 \dots, - 5.29252353 \dots$